tensorflow-二分法求解一元方程
阿新 • • 發佈:2018-12-15
tensorflow程式設計還是比較麻煩~
#!/usr/bin/env python2 # -*- coding: utf-8 -*- """ Created on Mon Jul 24 08:25:41 2017 f(x)=x^3+2*(x^2)-45=0 二分法求解一元方程 @author: [email protected] """ import tensorflow as tf def fp(x): return tf.subtract(tf.add(tf.pow(x,3),tf.multiply(tf.pow(x,2),2.)),45.) i=tf.Variable(1,dtype=tf.int32) MIN_RESULT=-10. MAX_RESULT=10. a=tf.constant(MIN_RESULT,dtype=tf.float32) b=tf.constant(MAX_RESULT,dtype=tf.float32) n=tf.constant(300) result=tf.Variable([],dtype=tf.float32) x_result=tf.Variable(MIN_RESULT-1,dtype=tf.float32) fa_result=tf.Variable(fp(MIN_RESULT),dtype=tf.float32) TOL=1e-6 #本部落格所有內容是原創,如果轉載請註明來源http://blog.csdn.net/myhaspl/ def fp_cond(i,n,a,b,result,fa_result,x_result): return tf.logical_and(tf.less(i,n),tf.less(x_result,MIN_RESULT)) def fp_body(i,n,a,b,result,fa_result,x_result): p=tf.add(a,tf.divide((b-a),2.)) fp_result=fp(p) x_result=tf.cond(tf.logical_or(tf.equal(fp_result,0),tf.less((b-a)/2.,TOL)),lambda:p,lambda:MIN_RESULT-1) result=tf.concat([[a,b,x_result],result],axis=0) [a,b,fa_result]=tf.cond(tf.greater(tf.multiply(fa_result,fp_result),0),\ lambda:[p,b,fp_result],\ lambda:[a,p,fa_result]) i=i+1 return i,n,a,b,result,fa_result,x_result init_assign = tf.global_variables_initializer() i,n,a,b,result,fa_result,x_result=tf.while_loop(fp_cond,fp_body,\ loop_vars=[i,n,a,b,result,fa_result,x_result],\ shape_invariants=[i.get_shape(),n.get_shape(),a.get_shape(),b.get_shape(),tf.TensorShape([None]),fa_result.get_shape(),x_result.get_shape()]) with tf.Session() as sess: sess.run(init_assign) sess.run([i,n,a,b,result,fa_result,x_result]) print "在",sess.run(result[0]),"-",sess.run(result[1]),"內:" print sess.run(i),"次迭代,計算方程的解:",sess.run(result[2])